Is there an entire function satisfying $|f(z)|=|z|+1$ for every $|z|\geq2017$? Is there a holomorphic function $f \in Hol(\Bbb C)$ such that $|f(z)|=|z|+1$ for every $|z|\geq2017$ ?
I tried defining $g(z)=\frac{1}{f(z)}$. Such $g$ is clearly holomorphic and bounded for $|z| \geq 2017$. I want to show that $g$ is holomorphic in $|z| \leq 2017$ and then conclude that $g$ is constant. For that I want to show that $f$ has no zeros in $|z| \leq 2017$. I tried to prove it by Rouche theorem, but didn't succeed. Any ideas?
Edit: the post similar to mine doesn't solve my problem, because the ending of the proof is not clear. Where is the contradiction?
 A: Partial answer. An entire function has a Taylor expansion
$$f(z)=a_0+\sum\limits_{n=1}a_nz^n \tag{1}$$
where 
$$a_n=\frac{f^{(n)}(0)}{n!}=\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{f(z)}{z^{n+1}}dz$$
As a result
$$|a_n|=\left|\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{f(z)}{z^{n+1}}dz\right|\leq
\frac{1}{2\pi}\int\limits_{|z|=r} \left|\frac{f(z)}{z^{n+1}}\right||dz|= \\
\frac{1}{2\pi}\int\limits_{|z|=r} \frac{|z|+1}{|z|^{n+1}}|dz|=
\frac{1}{2\pi}\int\limits_{|z|=r} \frac{r+1}{r^{n+1}}|dz|=\\
\frac{1}{2\pi}\frac{r+1}{r^{n+1}}\cdot 2\pi r=
\frac{r+1}{r^{n}}$$
Taking the limit $\lim\limits_{r\rightarrow\infty}$, we have that $|a_n|=0$ for all $n\geq2$, thus from $(1)$
$$f(z)=a_0+a_1z, |a_1|\leq 1 \tag{2}$$
Going further, it's easy to show that $|a_1|=1$ from 
$$|f(x)|=|a_0+a_1z|=|z|+1 \Rightarrow \left|\frac{a_0}{z}+a_1\right| = 1 + \frac{1}{|z|}$$
and taking $\lim\limits_{z\rightarrow\infty}$.
Can you continue with this?
A: Partial answer (continuation of rtybase's answer where we got to $f(z)=a_0+a_1z$).
Consider the circle $|z|=C$ where $C$ is a constant greater than $2017$. When $z$ runs over this circle, $a_1z$ runs over the circle centered at the origin with radius $|a_1|C$. 
Thus $f(z)$ runs over the circle centered at $a_0$ with radius $|a_1|C$. 
Your condition that $|f(z)|=C+1$ implies then that $a_0=0$ (the centre of the circle where $f(z)$ lives), and more importantly that $|a_1|C=C+1$ for all $C>2017$. This last condition is impossible since $\frac{C+1}{C}$ is not a constant.
